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• 1997 AIME Problems/Problem 6
  Using [[Simon's Favorite Factoring Trick|SFFT]], {{AIME box|year=1997|num-b=5|num-a=7}}
  3 KB (497 words) - 00:39, 22 December 2018
• 1997 AIME Problems/Problem 3
  ...1000x+y=9xy \Longrightarrow 9xy-1000x-y=0</math>. Using [[Simon's Favorite Factoring Trick|SFFT]], this factorizes to <math>(9x-1)\left(y-\dfrac{1000}{9}\right) {{AIME box|year=1997|num-b=2|num-a=4}}
  2 KB (375 words) - 19:34, 4 August 2021
• 1998 AIME Problems/Problem 3
  Factoring the first one: (alternatively, it is also possible to [[completing the squa {{AIME box|year=1998|num-b=2|num-a=4}}
  1 KB (198 words) - 20:13, 23 February 2018
• 2000 AIME I Problems/Problem 9
  ...<math>\log x, \log y, \log z</math> respectively. Using [[Simon's Favorite Factoring Trick|SFFT]], the above equations become (*) Equating the first and second equations, solving, and factoring, we get <math>a(1-b) = c(1-b) \implies{a = c}</math>. Plugging this result
  4 KB (623 words) - 15:56, 8 May 2021
• 2000 AIME I Problems/Problem 7
  ...</math> and <math>\hspace{0.05cm}\frac{1}{z} = xy</math>. Substituting and factoring, we get <math>x(y+1) = 5</math>, <math>\hspace{0.15cm}y(z+1) = 29</math>, a {{AIME box|year=2000|n=I|num-b=6|num-a=8}}
  5 KB (781 words) - 15:02, 20 April 2024
• 2000 AIME I Problems/Problem 5
  Factoring <math>b_1 \cdot b_2 = 54</math> and saying WLOG that <math>b_1 < b_2 < 25< {{AIME box|year=2000|n=I|num-b=4|num-a=6}}
  7 KB (1,011 words) - 20:09, 4 January 2024
• 2003 AIME I Problems/Problem 8
  We then set <math>400+x^2=y^2</math>, where <math>y</math> is an integer. Factoring using difference of squares, we have ...<math>d</math>, and this must be the only possible answer since this is an AIME problem. We got very lucky in this sense :)
  5 KB (921 words) - 23:21, 22 January 2023
• 2001 AIME II Problems/Problem 6
  ...tion, <math>(b - a/2)^2 + (2b - a - a/2)^2 = a^2/2</math>. Simplifying and factoring, we get <math>2a^2-7ab+5b^2 = (2a-5b)(a-b) = 0.</math> Since <math>a = b</m {{AIME box|year=2001|n=II|num-b=5|num-a=7}}
  4 KB (772 words) - 19:31, 6 December 2023
• 2000 AIME II Problems/Problem 13
  Factoring a bit, we get {{AIME box|year=2000|n=II|num-b=12|num-a=14}}
  6 KB (1,060 words) - 17:36, 26 April 2024
• Mock AIME 1 2006-2007 Problems/Problem 9
  ...ual to <math>2^{10}\binom{10}{0}y^{10}(-1)^0</math> and by simplifying and factoring , we get <math>y^{10}(1024-1)=y^{10}(1023)</math>. Thus, <math>1023</math> *[[Mock AIME 1 2006-2007 Problems/Problem 8 | Previous Problem]]
  5 KB (744 words) - 19:46, 20 October 2020
• Mock AIME 2 2006-2007 Problems/Problem 3
  ...so <math>(n+6)^2 = m^2 + 2043</math>. Subtracting <math>m^2</math> and [[factoring]] the left-hand side, we get <math>(n + m + 6)(n - m + 6) = 2043</math>. < {{Mock AIME box|year=2006-2007|n=2|num-b=2|num-a=4}}
  1 KB (198 words) - 10:50, 4 April 2012
• Mock AIME 3 Pre 2005 Problems/Problem 15
  Factoring the radicand, we have {{Mock AIME box|year=Pre 2005|n=3|num-b=14|after=Last Question}}
  2 KB (312 words) - 10:38, 4 April 2012
• Sophie Germain Identity
  ...out the right side and verifying that it equals the left. To derive the [[factoring]], we begin by [[completing the square]] and then factor as a [[difference ...^4+324)(40^4+324)(52^4+324)}</math>. ([[1987 AIME Problems/Problem 14|1987 AIME, #14]])
  2 KB (222 words) - 15:04, 30 December 2023
• Algebraic manipulation
  ...te perfectly valid solutions. Thus, when possible, it's often better to [[factoring|factor]] than to divide both sides. ...anipulation to get all terms to one side to form a [[quadratic]]. After [[factoring]], we get <math>x = 5, -3</math>.
  4 KB (562 words) - 18:49, 8 November 2020
• Mock AIME 1 Pre 2005 Problems/Problem 5
  ...]{20-x}</math>. Then <math>a+b = 2</math> and <math>a^3 + b^3 = 20</math>. Factoring, <cmath>a^3 + b^3 = (a+b)((a+b)^2-3ab) = 2(4-3ab)= 8-6ab=20 \Longrightarrow {{Mock AIME box|year=Pre 2005|n=1|num-b=4|num-a=6|source=14769}}
  979 bytes (159 words) - 17:46, 21 March 2008
• 2008 AIME I Problems/Problem 13
  Plugging this into the first equation, and factoring, and cancelling <math>(x+1)</math>, and simplifying, we get <math>19x^2 -43 {{AIME box|year=2008|n=I|num-b=12|num-a=14}}
  8 KB (1,218 words) - 00:07, 11 April 2024
• 2008 AIME II Problems/Problem 1
  Factoring this expression yields {{AIME box|year=2008|n=II|before=First Question|num-a=2}}
  4 KB (575 words) - 16:41, 14 April 2024
• 2010 AIME I Problems/Problem 15
  ...s, corresponding to the sides of the triangle and thus degenerate cevians. Factoring those out, we get <math>0 = 4x(x - 15)(3x^2 - 40x + 132) = x(x - 15)(x - 6) {{AIME box|year=2010|num-b=14|after=Last Problem|n=I}}
  14 KB (2,210 words) - 13:14, 11 January 2024
• 2021 AIME II Problems/Problem 3
  ==Solution 5 (Factoring)== {{AIME box|year=2021|n=II|num-b=2|num-a=4}}
  7 KB (1,167 words) - 03:52, 11 March 2023
• 2021 AIME II Problems/Problem 6
  ...|A|+|B|-|A \cap B| = |A \cup B|</math>. Substituting into the equation and factoring, we get that <math>(|A| - |A \cap B|)(|B| - |A \cap B|) = 0</math>, so ther Rearranging and applying Simon's Favorite Factoring Trick give
  8 KB (1,364 words) - 01:02, 29 January 2024

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